Answer
\[
\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| = 1
\]
Solution
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
\(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(-1\right)^{n+1}}{\left(n+2\right)\left(n+4\right)}\times \frac{\left(n+1\right)\left(n+3\right)}{\left(-1\right)^{n}}\right|\right)\)
- step1: Multiply the terms:
\(\lim _{n\rightarrow +\infty}\left(\left|-\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\)
- step2: Calculate the absolute value:
\(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\)
- step3: Rewrite the expression:
\(|\lim _{n\rightarrow +\infty}\left(\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right)|\)
- step4: Calculate:
\(|1|\)
- step5: Calculate the absolute value:
\(1\)
Calculate the limit \( \lim_{n \rightarrow \infty}\left|\frac{(-1)^{n+1}}{(n+2)(n+4)} \cdot \frac{(n+1)(n+3)}{(-1)^{n}}\right| \).
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
\(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(-1\right)^{n+1}}{\left(n+2\right)\left(n+4\right)}\times \frac{\left(n+1\right)\left(n+3\right)}{\left(-1\right)^{n}}\right|\right)\)
- step1: Multiply the terms:
\(\lim _{n\rightarrow +\infty}\left(\left|-\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\)
- step2: Calculate the absolute value:
\(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\)
- step3: Rewrite the expression:
\(|\lim _{n\rightarrow +\infty}\left(\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right)|\)
- step4: Calculate:
\(|1|\)
- step5: Calculate the absolute value:
\(1\)
To find \( \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \), we start by substituting the expression for \( a_n \):
\[
a_n = \frac{(-1)^{n}}{(n+1)(n+3)}
\]
Now, we calculate \( a_{n+1} \):
\[
a_{n+1} = \frac{(-1)^{n+1}}{(n+2)(n+4)}
\]
Next, we find the ratio \( \frac{a_{n+1}}{a_n} \):
\[
\frac{a_{n+1}}{a_n} = \frac{\frac{(-1)^{n+1}}{(n+2)(n+4)}}{\frac{(-1)^{n}}{(n+1)(n+3)}} = \frac{(-1)^{n+1}}{(-1)^{n}} \cdot \frac{(n+1)(n+3)}{(n+2)(n+4)} = -\frac{(n+1)(n+3)}{(n+2)(n+4)}
\]
Taking the absolute value:
\[
\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)(n+3)}{(n+2)(n+4)}
\]
Now, we compute the limit as \( n \) approaches infinity:
\[
\lim_{n \rightarrow \infty} \frac{(n+1)(n+3)}{(n+2)(n+4)}
\]
Dividing the numerator and the denominator by \( n^2 \):
\[
= \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{\frac{2}{n} + \frac{4}{n^2}} = \frac{1 \cdot 1}{1 \cdot 1} = 1
\]
Thus, we have:
\[
\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| = 1
\]
So the final answer is:
\[
\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| = 1
\]
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